If the real part of the complex number $${(1 - \cos \theta + 2i\sin \theta )^{ - 1}}$$ is $${1 \over 5}$$ for $$\theta \in (0,\pi )$$, then the value of the integral $$\int_0^\theta {\sin x} dx$$ is equal to:

Let $$f:R - \left\{ {{\alpha \over 6}} \right\} \to R$$ be defined by $$f(x) = {{5x + 3} \over {6x - \alpha }}$$. Then the value of $$\alpha$$ for which (fof)(x) = x, for all $$x \in R - \left\{ {{\alpha \over 6}} \right\}$$, is :

If $$f:R \to R$$ is given by $$f(x) = x + 1$$, then the value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {f(0) + f\left( {{5 \over n}} \right) + f\left( {{{10} \over n}} \right) + ...... + f\left( {{{5(n - 1)} \over n}} \right)} \right]$$ is :

Let A, B and C be three events such that the probability that exactly one of A and B occurs is (1 $$-$$ k), the probability that exactly one of B and C occurs is (1 $$-$$ 2k), the probability that exactly one of C and A occurs is (1 $$-$$ k) and the probability of all A, B and C occur simultaneously is k², where 0 < k < 1. Then the probability that at least one of A, B and C occur is :